
..MIS' 



Conservation Resources 
Lig-Free® Type I 
Ph 8.5, Buffered 



Tfl 683 
.M15 
Copy 1 



Fundamentals of 

Reinforced Concrete 

Design 



A lecture prepared by Ernest 
McCullough, Chief Engineer, Fireproof 
Construction Bureau, Portland Cement 
Association, for the Short Course for 
Manual Training and Vocational 
Teachers, held at Lewis Institute, 
Chicago, June 26 to July 1, 1916 



'Concrete for Permanence'* 



Published by 

Portland Cement Association 

111 West Washington Street 
Chicago 



October, 1916 



ATLANTA NEW YORK 

DALLAS PARKERSBURG. W. VA. 

INDIANAPOLIS PITTSBURGH 

KANSAS CITY SAN FRANCISCO 



Thomas A. Edison says: 

"The millions of dollars of fire 
losses in this country annually 
make it a matter of moment that 
the superiority of reinforced 
concrete for fireproof structures 
should be thoroughly understood 






FUNDAMENTALS 

of 

REINFORCED CONCRETE DESIGN 

Reinforced concrete is a combination of concrete and metal, preferably 
steel, with the two materials so disposed as regards position and amounts 
that each resists the stresses it is best fitted to resist. In piers, posts 
and columns the concrete takes compression assisted by the steel, and 
the vertical steel takes tension if any bending occurs. In beams three 
stresses act; namely, compression, tension and shear. The concrete 
takes all the compression and a limited amount of shear. The steel is 
computed as taking all the direct tension and assists the concrete to carry 
shear. 

To fully understand the principles of reinforced concrete it is best to 
first consider materials uniform in composition, such as wood, iron or 
steel. Take for example a square wooden post having a length less than 
15 times the thickness. This length is chosen because after the length 
exceeds 15 times the thickness bending can occur under heavy loading. 
The ratio of length to thickness (the slenderness ratio) then becomes a 
factor in the rules and formulas. 

The column chosen is 8 x 8 in. The length is immaterial so long as 
we have it less than 8X15-^12 = 10 feet. The cross-sectional area is 
8X8 = 64 sq. in. The allowable unit load or compressive stress for the 
wood considered is 800 lb. per sq. in. The column can safely sustain a 
load of 8X8X800 = 51,200 lb. 

B" 



00 




yerf/ca/ 

reinforcement' 



Fy-t 



Let us now consider a reinforced concrete column. For the benefit 
to be obtained by comparison we will make it also 8 x 8 in. outside dimen- 
sions. The steel will be in the form of four bars each 3^ in. square. 

The bars are set in the corners, as shown in Fig. 1, and lj^ in. in from 
the sides of the column. This is necessary for fire protection. The 
concrete outside of the steel (used for fire protection) assists in carrying 



/7-//W 



the load until the fire comes but after a severe fire it should not be depended 
on so we neglect it entirely in our computations. The actual area of 
the column is therefore 5X5=25 sq. in. Four Y2 m - square bars have 
an area of one square inch. The ratio of steel to concrete is 1 -=- 25 = . 04. 
The steel ratio multiplied by 100 is the per cent of steel reinforcement. 
Our column therefore contains 4 per cent of steel, the maximum for a 
column of this type. 

Some building ordinances limit it to 3 per cent, following the lead of 
Chicago. 

When a load is applied to the top of a column and the steel bars get 
their share they may bend because the slenderness ratio is large. It is 
necessary to put ties around the upright bars and these ties are spaced 
at intervals not exceeding 12 times the thickness of the vertical bars. 
Therefore, in the column under consideration, the ties will be spaced 
6 inches apart because the bars are only Y2 mcn square. The ties are 
held in place by No. 18 black stove wire and the ends are turned in far 
enough to be gripped by the concrete so they cannot be pulled out when 
stressed. Ties are usually made of heavy wire or J^ or J^-inch round 
steel rods. 

Having arranged the bars and the ties, how much will our column 
carry? 

Let us assume a 1:2:4 concrete with an allowable fiber stress in 
compression of 400 lb. per sq. in. The area of the concrete is 25 sq. in. 
less 1 sq. in. of steel = 24 sq. in., which at 400 lb. gives 24X400 = 9,600 lb. 
To determine the strength added by the steel we must be governed by 
the ratio of deformation between the steel and concrete. This, for the 
concrete we are using, is 15, as determined by experiments. 

Assume a piece of steel fastened in a vertical position and a load 
placed on top. Assume a piece of concrete of the same size similarly 
placed and loaded with an equal load. Careful measurements will show 
that both materials shorten under the applied loads but the decrease in 
the length of the steel is 1/15 that of the concrete. To produce equal 
shortening (deformation) under equal loads the cross-sectional area of 
the concrete must be 15 times the cross sectional area of the steel. Thus 
each square inch of steel is equal to 15 square inches of concrete. 

Now apply this to the column in question. The area of concrete is 
24 square inches. The area of the steel is 1 square inch, the equivalent 
of 15 square inches of concrete. Consider the area of concrete to be 
increased, making it 24 + 15 = 39 sq. in. The load-carrying capacity of 
the column is now 39X400 = 15,600 pounds. The average stress is 
15,600 

=624 lb. per sq. in., an increase of 5Q per cent. The unit stress 

25 
on the steel is 15X400 = 6,000 lb. per sq. in. 

A safe compressive stress for the steel alone would be 12,000 pounds 
per square inch, which shows that it is not economical to use steel in 
compression in reinforced concrete, except in columns. 

We cannot use a steel stress exceeding the concrete stress multiplied 
by the ratio of deformation or the concrete will be stripped from the steel 
and the column will fail. The two materials must act together and 
shorten equally, each carrying a proportion of the load. The ratio of 
deformation is therefore a stress ratio for columns or for members acting 
wholly in compression. 



The following formulas are used for the design of columns in which 
the unsupported length does not exceed 15 times the effective diameter 
or thickness; that is, the thickness of the column after deducting the 
protective covering of the steel. 

Let/ = average unit stress per sq. in. of effective area. 

/ c = allowable unit stress per sq. in. on plain concrete. 

p = ratio of steel to concrete. 

A c =area of concrete in sq. in. 

A s = area of steel in sq. in. 

A = total effective area = ^4 c +^4 s . 

n = ratio of deformation. 

P = total load. 
then P=Af=f c (A c +nA s ). 
or/=/ c [{l-p)+np\. 

THE HOOPED COLUMN 

Make a cylinder of thin paper and fill it with sand. The paper is 
barely strong enough to hold the sand and if a load is put on top the paper 
will burst and the sand will flow. Use a tin cylinder and the pressure 
required to burst it will be very great. Instead of sand use cement mortar 
or concrete and the metal casing can be made very thin, so thin, in fact, 
that a wire wound spirally around the concrete cylinder will furnish the 
necessary strength provided the amount of metal in the wire is equal to 
the amount found to be necessary in the solid thin shell. Poorly made 
concrete needs more reinforcement than first-class concrete. 

The hooped column consists of a concrete core reinforced with vertical 
steel and having a steel spiral around the core. There should be not less 
than eight vertical rods not exceeding 6 per cent of the area. The spiral 
hooping should be not less than one-half of one per cent and not to exceed 
one and one-half per cent of the area. More than this amount is wasteful, 
for it adds little strength. The spiral does not act until the concrete 
begins to fail and as it postpones the total failure the effect is the same 
as increasing the strength of the concrete in compression so we can use 
20 to 25 per cent higher unit stress, depending upon the building ordinance 
followed. Steel in the form of a spiral, provided it has a pitch not exceed- 
ing one sixth of the diameter, is 2.4 times as effective as the same amount 
placed vertically. The vertical equivalent of spiral steel is found as 
follows : 

Let c = circumference of the core in inches. 
x = pitch of spiral in inches. 
a = cross sectional area of steel used for spiral. 
A = Area of core in sq. in. 

f ca 
Then the equivalent ratio of spiral per foot of length =l.l{ 



v Ax J 
The strength of the hooped column is 

in which ^4 h = area of spiral steel, in terms of vertical steel. 

P 
or/= — =f c [{l-p)+np+<2.J h np'] 

A 
in which p' = ratio spiral steel expressed as equivalent vertical steel. 



BEAMS 

In Fig. 2 is shown a beam bending under load. In the middle of the 
span is shown a vertical line A c, an extension of a radial line. On one 
side of this line is a radial intercept A'b and on the other side a radial 
intercept A"d. 




bed 

Fig. 2 

Provided the material is homogeneous, that is, uniform in quality 
and strength, and is not stressed, beyond the elastic limit, a vertical section 
plane before the beam bends is plane after it bends. That is, A c is straight 
before the load is applied and the lines A'b and A"d are also straight 
although the horizontal separation be is greater than A A' and cd is greater 
than A A" . In Fig. 3 the line A'b is assumed to be moved across Ac 
so the space A A' = b'c. This is equivalent to revolving the line Ac until 
it becomes A' b', parallel to A'b. 

In Fig. 4 this is again shown to illustrate the two force triangles, 
the upper one representing compression and the lower one tension. The 



A' A A' 


I 
i 
i 
I 

* 
- — ~ _ _ * 

Tl 


r 
I 

l 

f 

I 



bb'c 

Fig. 3 




material being homogeneous the neutral axis x x' is midway be- 
tween the top and bottom edges. The force triangles are therefore equal, 
the stress being zero at the neutral axis and a maximum at the edges. 
The maximum unit stress (skin stress some men call it) is designated by 

/ 

the letter /. The average stress is — . The area of each force triangle is 

/ h Jh. 

2 2k 

We have been considering a thin slice of a beam, and as a beam has 
breadth we will use the letter b (breadth) to designate this. Our force 

fkb 

triangles now become wedges each with a volume = — — 

h 

Forces act through the center of gravity of bodies and the center of 

h 
gravity of a triangle is — from the base. The distance between the 

3 

2h 
center of gravity of the two force triangles is — as shown in Fig. 4. 

3 
The total compressive force is equal to the total tensile force exerted 
to resist bending and each force wedge acts with a moment arm 

2h 
— — , so we obtain the moment of resistance by multiplication, thus 
3 
2h fbh 2fbh 2 fbh 2 

M = — X — — = for a rectangular beam of homogeneous material; 

3 k 12 6 
that is, one in which (below the elastic limit) the tensile strength equals 
the compressive strength. 

In Fig. 5 is shown, a beam made of two pieces with a hinged joint. 
In the top of the joint is a block of rubber and at the bottom is a coiled 
spring. When a load is placed on top the beam will, of course, bend at 



f — ffubber 




u^ C of tec/ epring 
£, 3uppart ftp. S Support^ 

thelhinged joint. It requires no effort of imagination to prove that in 
the|top of the open joint the tendency to close is opposed by the rubber 
and in the bottom the tendency to open is opposed by the spring. Actually 



the hinge midway is not required. It merely locates definitely the position 
of the neutral axis, and to consider the hinge as a necessary feature is 
likely to confuse one as to the action of resisting forces in a beam. The 
neutral axis is the point where the character of stress changes from tension 
to compression, or from compression to tension. In a beam of homo- 
geneous material, that is, one in which the tensile and compressive strengths 
are equal, with symmetrical cross-section, the neutral axis will be midway 
between the top and bottom surface, or skin. At the skin the stress is 
a maximum. At the neutral axis it is zero. A diagram illustrating this 
is triangular and is termed a force triangle. 

In Fig. 6 (a) the compressive triangle has vertical lines and the tensile 
triangle has horizontal lines. Each triangle overlaps the other and the 
heavily shaded diamond center indicates a cancellation of one force by 
another. 

The remaining effective stresses are shown in Fig. 6 (b). The neutral 
axis is therefore the point where the tensile and compressive stresses are 
definitely separated. In a beam with a finite breadth, for we have been 
considering only a thin vertical slice, the neutral axis becomes the neutral 
plane. The use of the word neutral implies a point, place or plane where 
there is a definite neutralization of opposite forces, or stresses. The 
force acting along the neutral plane is therefore horizontal shear, for the 
forces acting on either side are opposite in character and equal in magni- 
tude. 





<*) 



ft) 



Fig. 6 



The principle of the lever is evident. The length of the lever arm 
is the distance between centers of gravity of the opposite forces and the 
fulcrum is situated in the neutral plane. 

In a reinforced concrete beam steel is placed near the lower edge to 
take all the tension, for, roughly speaking, concrete is ten times stronger 
in compression than in tension. In a plain concrete beam the neutral 
plane will be very high because the tensile and compressive forces must 
be equal. The tensile stress will be low and the compressive stress will 
be high, but the lever arm is a constant. The relative volumes of the two 
force wedges will be approximately as 10 is to 1. 

When steel reinforcement is used the area required is computed on 
the assumption that it will carry all the tension and the value of the con- 
crete in tension below the neutral axis is neglected. The tensile stress 
therefore does not vary from zero at the neutral axis to a maximum, 
but the compressive stress above the neutral axis does so vary. 

The ratio of deformation between concrete and steel prevents the 
consideration of the value of concrete in tension. Experiments have 
shown that when steel embedded in concrete is stressed in tension to an 
amount practically equal to the tensile strength of an area of concrete 
equal to the steel area, multiplied by the ratio of deformation, the concrete 



cracks. The tensile strength per sq. in. in the concrete at the level of 

/• . . 

the steel = — , in which L = unit tensile steel stress. 



n 



PB These cracks are vertical and fairly uniformly spaced. They probably 
extend as far into the beam as the neutral axis when the beam is on the 
point of failure. If the beam is well made and the bond of the concrete 
to the steel is good the cracks are so small, because numerous, that there 
is no danger of the entrance of moisture in large enough amounts to cause 
rusting of the steel. It is therefore possible to use steel with a very high 
stress, for the ratio of deformation is not a stress ratio as in the case of 
columns carrying direct compressive stress. In a column a high com- 
pressive stress on the concrete may strip it from the steel. In beams a 
high tensile stress in the concrete merely cracks it and the concrete between 
the cracks clings to the steel and protects it from corrosion. The vertical 
tension cracks in beams are shown in Fig. 7. 





-i-H-HV . 

Fig. 7 

The ratio of deformation plays an important part in determining 
the location of the neutral axis in concrete beams. In the Chicago 
Building Code the following values are used as a result of experiments: 



Mixture 


1 


1 :£ 


1 


1J"2 :3 


1 


2 :4 


1 


2^:5 


1 


3 :7 



• Ultimate Compressive 


Ratio of 


Strength Per Square Inch 


Deformation 


2,900 


10 


2,400 


12 


2,000 


15 


1,750 


18 


1,500 


20 



The allowable safe unit stress per sq. in. is thirty-five hundredths of 
the ultimate strength in compression. 

In reinforced concrete we have two materials with widely differing 
unit stresses. The letter / is used to denote the unit stress per square 
inch, usually termed the "Fiber Stress." The unit steel stress is / 8 and 

the unit concrete stress is / c . The stress ratio — is denoted by the letter 

/. 

m (meaning "measure"). The ratio of deformation is denoted by the 
letter n (meaning "number"), for it is an arbitrary number which is 
approximately correct. 

Fig. 8 is a graphical representation of the effect n and m have on the 
location of the neutral plane in a reinforced concrete beam. On a piece 
of quadrille ruled paper plot the depth from the top of the beam to the 



center of gravity of the steel by setting off ten divisions. Set off on the 
same scale the ratio of stresses (m) and the ratio of deformation (n) as 
shown. The depth to the neutral axis, fc, may then be scaled. The exact 



n 



value is k = ■ 



n-\-m 





1 


• "1 


l 


T "-., S 




k --*■* ^ 










d* 


i ^^ """^ 


/ ^ "^v 




• .? ^< 




/-/C ^ ""-^ 




1 ^ ^i*. 




' 1 x^ ^_ 




AT7 a ftfj 





A> 










t 

♦ 


r*; 






/77 

* »> 



Example. What is the value of k when / s = 16,000 lb. per sq. in. and 
fc = 650 lb. per sq. in.? 
n = 15 
16000 

= 24.62 



m 



k = 



650 

71 



15 



■ = 0.878 d 



n+m 15+24.62 

Fig. 9 shows the force triangle of the concrete in compression and the 
steel in tension. To find the ratio of steel, proceed as follows: 

The total amount of compressive force is found by obtaining the area 
of the force triangle. The height is kd and the average stress is 



kef 

■+- 

■■rc4r=: 



*\ 




Neutral Axis 



M 



JtX- 



Steel 



/Vf.S 



— The compression = . We use kd for the depth, d, has a definite 

2 2 

value ana k is a percentage of d. The moment arm, the lever, has a 
kd k 

length = d or jd.(j = l ). The total compressive force in a thin 

3 3 

fc fc 

slice, C — — kdjd = — kjd 2 

2 2 



10 



The tensile force T=fsAjd, in which ^4 = area of steel in sq. in. The 
A 
steel ratio = — = p, so it is necessary to introduce the breadth b, into 
bd 



the expression. 
We now have 






C-- 


/« 

= —kjbd 2 

2 










and 


















then 






T- 


=fspjbd 2 










/«* 

— =f s p, provided C- 
2 
reinforcement. 


= T, 


which is 


the case 


for 


a beam 


with 


"balanced" 




















650 
















Assigning values, 
and 


2 


xo 


.378 = 


= 16000p 












325 X. 


378 












P = 






= ( 


). 00767 











16000 
pX100 = per cent of steel = . 767 (0 . 77%) . 
k 

By formula p = 

2m 

Assume a beam 6 inches wide with depth to the center of the steel = 
9 ins. What is the resisting moment? 

C = 325X0. 378X0. 874 X6X9 2 = 52,242 in lb. 

What area of steel will be required? 

4 = 6dp = 6X9X0. 0077 = 0. 42 sq. in. 

Check the steel: 

r=/^jd = 16000X0.42X0.874X9 = 52,859inlb. 

The greater resisting moment in tension is due to having used 0.42 
sq. in. of steel, the exact area being 0.4158 sq. in. The area of steel used 
is governed by the commercial sizes of bars and rods, or the expanded 
metal or wire fabric used. The actual steel area used will usually be 
greater than the theoretical area necessary. 

Every reinforced concrete beam has two moments of resistance; one 
determined by the concrete, the other by the steel. The lesser of the two 
is the resisting moment which determines the actual strength. In design- 
ing slabs a width of one foot is taken, for a slab is merely a wide and 
shallow beam assumed to be made up of a number of beams each 12 
inches wide. 

fjcj 

R is a moment factor. For the concrete R = . For the steel 

2 
R = f s pj. Then the bending moment M = Rbd 2 

M 

b = 

Rd 2 



/M 
d=V — 

Rb 

li 



To design a beam select stresses for the steel and concrete and find R. 
The value of M is the bending moment which is equal to, or is less than, 
the resisting moment. Assume a breadth and solve for the depth, or 
assume a depth and solve for the breadth. T beams are beams in which 
the floor slab is considered to be a part of the beam and carries the com- 
pression. The breadth, b, is the width in the floor slab and the stem 
below the slab must be wide enough to contain the reinforcement. The 
width necessary must include space between the bars and on each side 
to furnish bond and shearing strength. 

Below the steel there must be a covering of concrete not less than 
the thickness of the steel, with a minimum thickness of one-half inch, 
this being for bond and fire protection. In all building ordinances mini- 
mum coverings of concrete are specified; as for example, one-half inch 
for slabs and one and one-half inches for beams, girders and columns. 

SHEAR 

Fig. 10 shows a beam with typical shear cracks. A beam may fail 
by crushing of the concrete in the top, by the stretching or slipping of 
the steel or by shear, which is manifested by the appearance of shearing 




cracks. These cracks are an indication of tension in the concrete and 
stirrups are used to prevent shearing (diagonal tension) failures. 

Fig. 11 (a) shows a uniformly loaded beam resting freely on two 
supports. At (b) is shown the shear diagram. The vertical shear at 
either end is equal to one half the load and is zero at the point of maximum 
bending moment. The vertical depth, measured in pounds, of the shear 
diagram at any point is a sum-curve of the loading to that point. (See 
page 13.) 

The bending moment at any point is the area of the shear diagram 
between that point and the support. The vertical dimension is in pounds. 
The horizontal dimension is in feet when the result is foot pounds and in 
inches when the result is inch pounds. The bending moment at any 

section, such as y y is the sum curve of the shear at that 

section. 

Let the bending moment at y y be M (in inch pounds) , 

M 
then the unit fiber stress at y y = . 

jdb 

Let V = total shear at any section. 

M = bending moment on one side of the section, and 
M' = bending moment on the other side of the section, the section 
assumed to have a thickness as nearly zero as possible. 

12 



Then on one side 



/= 



M 



And on the other side 
f= 



jdb 



W 



jdb 



The total shear is V = M — M f , and the unit shear is 

V M-M' 

v — = 

jdb jdb 



\*--Para£>o/a 




In T beams the b used in expressions for shear is the thickness of the 
stem below the floor slab. 

The shear being tension increases from the top and bottom skin to 
a maximum at the neutral axis where the bending stress changes from 
tension to compression and from compression to tension. In a beam of 
homogeneous material with a uniform cross section the shear is the same 
above and below the neutral plane at equal distances from this plane. 
The amount may be computed for any depth by assuming different values 
for jd, the V and M being constant. This distribution of horizontal 
shear is found in a reinforced concrete beam above the neutral plane. 



13 



Below the neutral plane the shear is constant, for all the tension is carried 
by the steel. The unit value of the shear at all depths from the neutral 
axis to the center of gravity of the steel is determined by the expression, 

V 



jdb 
in which d = depth from top of beam to center of gravitv of the steel 

h 

3 = 1 — 
3 
k 
jd = d — 
3 
In Fig. 12 is given an illustration of horizontal shear. Several planks 
laid loosely on end supports bend under load and the slipping of one plank 
past the adjoining plank is a horizontal movement. This is shown in 
Fig. 12a. Spike, or bolt, the planks together as shown in Fig. 12b and 
the slipping cannot occur. The spikes or bolts represent with fair 
accuracy the stirrups used in reinforced concrete beams. The shear 
being a maximum at the ends where the bending moment is a minimum, 
the fastenings are closer together than nearer the middle of the span 
where the shear is a minimum and the bending moment is a maximum. 




(a) 





nr— t 

i—i — r 



T— h 






T=a 



-! — L-4. 




i 



S^ 



By reference to Fig. lib it is seen that vertical shear exists at all 
points on a beam. We have seen that horizontal shear also exists at all 
points on a beam. Along the neutral plane it is equal to the maximum 
end vertical shear. Very thin horizontal slices, like the planks in Fig. 12, 
are assumed for the purposes of computation. 

The resultant, according to the parallelogram of forces, is a diagonal 
tension, which causes the cracks shown in Fig. 10. 

In designing reinforced slabs, beams and girders the resisting moment 
must be equal to or greater than the bending moment. When this is 
fixed the beam must be tested for shearing strength, and if it is found to 
be deficient in this particular, steel in the form of stirrups must be provided 



14 



or the size of the beam be increased. The unit shearing stress v should 
not exceed 40 lb. per sq. in. for the concrete alone, nor exceed a total of 
120 lb. per sq. in. when stirrups are used. 

BOND 

Experiments have shown that 70 lb. per sq. in. is a safe allowance 
for bond in order that the steel and concrete may act together. The 
coefficient of expansion of the two materials is practically the same, so 
we need not fear a separation under extreme variations in temperature. 

Assuming the two materials to act together and the safe bond stress 
is 70 lb. per sq. in., what length of embedment is necessary for a bar one- 
inch square stressed 16,000 lb. per sq. in.? 

The one-inch square bar has four square inches of surface for each 
inch of length. 

16,000 

= 57.142 in. embedment required. 

4X70 

Four 3/2-inch square bars have the same area as one 1-inch square bar 
but the surface = 4 (4XH) = 8 sq. in. and the length of embedment 

16,000 
= = 28. 57 in. 

8X70 

When a large bar having sufficient area to carry the tension is found 
to be deficient in bonding area, smaller bars may be used. The stress is 
the maximum tensile stress in the reinforcement at the point of maximum 
bending moment, and the reinforcement each side of this point must be 
long enough for bond. 

BENDING UP STEEL 

The bending moment decreases toward the supports and when a 
number of bars are used they may gradually be decreased in number, 
always allowing not less than two to go the full length in the bottom. 
The other bars are turned up a short distance past the point where they 
are no longer needed for direct tension, being carried to within an inch 
of the top of the beam on an angle of 45 degrees or less, and thence horizon- 
tally to the supports. Half the steel area may thus be bent up at 0.23 
of the span from the support, one half the remainder at 0.15 of the span 
and one half of the remainder at 0.10 of the span. These rules are closely 
approximate and apply only to uniformly loaded beams. Exact rules are 
given in text books. Bent up steel assists in reinforcing the web or body 
of the beam and thus strengthens the beam against failure by shear. 

STIRRUPS 

Stirrups should be fastened to the tension steel, not merely looped 
around it. They should extend far enough above the neutral axis to 
develop bond. The stress in the stirrups is tension and it is equal to the 
tension in the concrete multiplied by the ratio of deformation at the 
instant the concrete cracks, when the whole stress is immediately taken 
by the steel. The stress used for stirrups should not exceed 75 per cent 
of the stress in the tension reinforcement. 

15 



Fig. 13 represents a unit vertical slice of a beam. 

V = total vertical shear at any section. 

w = load in pounds per lineal foot. 

L = span of beam in feet. 

b = breadth of beam in inches. 

d = depth to center of steel in inches. 

j = moment arm (percentage of d) . 

# = unit shear. 

v c = unit shear carried by concrete. 

v s = unit shear carried by steel. 

/ = distance in inches to point where concrete alone can carry the 
shear. 
T' = total tension of horizontal shear due to shear carried by the steel. 
D t = diagonal tension at neutral plane. 
slf s = unit tensile steel stress. 
^4 w = area of web steel = ^4 b +$ a . 
A h = area of bent up bars. 
I iS a = area of stirrup steel. 
Then (for uniformly loaded beams) — 
wL wL 
(1) V — at supports = wx at any point distant, x, from support. 



2 



(See Fig. 11.) 
V 
(fi) v = (max. 1201b.) 



2 



jdb 

{Jp) I (in ins.) = 



6vJj 



(3) v s = v — v c (# c = 40lb. per sq. in.) 
LvJb 



(5) r 



2 



r 



(«)A=- 



0.707 T' (7) A„ = 



D t 



(8) S, = A W -A b . 



/. 




Fig. /3 



The stirrups should be equal in area and the intervals should increase 
by^ increments of 50% or logarithmically, the greatest interval being equal 
to d. The area of steel in the bent up bars should be neglected when it 
is more than 0.15L from the supports, for the maximum shear is at the 
edge of the support at either end. When four stirrups are used at each 
end, the first will be \^d from the support; the second }/^d from the first; 
the third %d from the second and the fourth d from the third. 



16 



REINFORCING A CIRCULAR TANK 

wd 

In a round tank the only stress is tension, being equal to . 

2 
in which w = weight (pressure) . 
d = diameter. 
The tension is figured for strips one foot wide, the weight used being 
the weight of the liquid at the depth of the strip. The weight of one 
cubic foot of water being 62.5 lb., w at the depth of 10-ft. is 625 lb. At 
15 feet the weight is 937.5 lb. 

The stress to be used in the steel cannot exceed — , or large cracks will 

n 
open. The concrete shell should be designed to carry all the stress with 
an assumed safe unit tensile stress. Steel should be used to carry all the 
tensile stress. This combination fixes the unit stress in the steel as equal 
to nf c until something causes the tank wall to crack, after which all the 
tension is carried by the steel with the assumed fiber stress, f s . 

Owing to the difficulties encountered in construction no tank wall 
should have a thickness of less than six inches, regardless of the theoretical 
thickness found by computation. 

Example: Give the proper thickness of wall and amount of steel 
required for a circular tank 20 feet in diameter at a depth of 14 feet, using 
a 1 : 2 : 3 concrete. 

wd 14X62.5X20 

T = = -8,750 lb. 

2 2 

Use a tensile stress of 12,000 lb. per sq. in. in the steel in order to care 
for possible mistakes in connecting ends of bars. Clamps should not be 
used. The best method is to have the ends of the bars overlap a length 
of not less than 40 times the thickness. The overlapping ends should 
not be in contact, but should be separated to leave a space of about twice 
the thickness of the bars so the concrete may surround the steel. 
8750 

A s = = 0.729 sq. in. (area of steel). 

12000 
Use three 3^2-in. sq. bars, giving an area of 0.75 sq. in. 
A safe tensile stress for well made 1:2:3 concrete is 175 lb. per sq. in. 
8750 

Area of concrete = = 50 sq. in. The theoretical thickness of the wall 

175 
50 
is — = 4.166 in., for the strip is 12-ins. wide. The thickness, for reason 

12 
given, should be not less than 6-ins., so the actual area will be 72 sq. in. 
The area of the concrete is 72 sq. in. minus area of steel = 72— .075 = 71.25 
sq. in. 

The ratio of deformation for 1 : 2 : 3 concrete is 12, so the steel is 
equivalent to a concrete area of 12X0.75 = 9 sq. in. 
Adding : 71.25 + 9 = 80.25 sq. in. 

The average stress is 8750-^72 = 121.53 lb. per sq. in., and the stress 
on the concrete is 8750^80.25 = 109.03 lb. per sq. in. 

17 



The stress in the steel, when both materials are carrying tension, is 
12X109.03 = 1308.36 lb. per sq. in. 

If, for any reason, the concrete cracks the steel will carry all the 
tension with a stress of 12,000 lb. per sq. in. Cracks may occur where an 
occasional poorly mixed batch of concrete was deposited; where con- 
struction joints are defective; where forms were removed too roughly; 
through ice pressure; by reason of excessive temperature changes; fre- 
quent and sudden changes in pressure due to quick filling or emptying 
of water; through defective foundations. When the steel stress is not 
permitted to exceed 12,000 lb. per sq. in. the only effect of cracking will 
be to throw the entire tension on the steel and the cracks will not be 
large enough to cause corrosion or leakage. 

LITERATURE 

This lecture is not intended to completely cover the subject of reinforced 
concrete design. It is merely an introduction to the subject. The follow- 
ing books are recommended as texts : 

"Principles of Reinforced Concrete Construction," by Turneaure & 
Maurer, $3.50, published by John Wiley & Sons, New York, N. Y. 

"Concrete, Plain and Reinforced," by Taylor & Thompson, $5.00, 
published by John Wiley & Sons, New York, N. Y. 

Three books by Prof. Geo. A. Hool, intended for self -tutored men; 
entitled "Reinforced Concrete Construction," published by McGraw-Hill 
Publishing Company, New York, N. Y. 

Vol. 1 Fundamental Principles, $2.50. 

Vol. 2 Retaining Walls and Buildings, $5. 

Vol. 3 Bridges and Culverts, $5. 

For data on practical design and construction, "Recommended Building 
Code," prepared by the National Board of Fire Underwriters, 76 William 
Street, New York, N. Y. This building code should be used in all technical 
schools as a text book on modern building construction. 



18 



THE EXTENSION DIVISION 
of the Portland Cement Association 
has, for the free use of schools, colleges or 
lecturers, the following sets of lantern slides : 

PROPORTIONING AND MIXING CONCRETE— 16 slides 

Selection and proportioning of concrete materials, hand 
mixing, depositing concrete. 

CONCRETE WALKS AND FLOORS— 24 slides 

Construction of concrete walks, feeding floors, barnyard 
pavements and miscellaneous work. 

CONCRETE TROUGHS AND TANKS— 34 slides 

Troughs, stock tanks, dipping vats, hog wallows, 
manure pits, cisterns and similar work are illustrated, 
with suggestions for their construction. 

CONCRETE FENCE POSTS— 26 slides 

Illustrations of concrete fence, line and gate posts and 
the means commonly employed in constructing them. 

PERMANENT FARM BUILDINGS— 34 slides 

The uses of concrete in constructing and improving 
barns, dairy buildings, hog and poultry houses and 
other farm structures. 

CONCRETE SILOS— 41 slides 

Various types of concrete silos and methods of con- 
struction. 

CONCRETE IN THE COUNTRY— 33 slides 

A general set of views showing numerous uses of con- 
crete on the farm and in the rural community. 



Write for descriptions, and conditions under which 
they may be obtained 




029 942 950 7 
Tennis Courts of Concrete 

Concrete Septic Tanks 

Concrete Fence Posts 

Small Concrete Garages 

Concrete Feeding Floors, Barnyard Pavements 

and Concrete Walks 
Specifications for Concrete Roads, Streets 

and Alleys 

Specifications for Concrete Pavement Between 
Street Car Tracks 

Fact s Every one Should Know About Concrete Roads 

Concrete Facts About Concrete Roads 

Proportioning Concrete Mixtures and Mixing 
and Placing Concrete 

Concrete Foundations 

Concrete Troughs, Tanks, Hog Wallows, Manure 
Pits and Cisterns 

Concrete Sewers 

Concrete Swimming and Wading Pools and 
How to Build Them 

Concrete Linings for Irrigation Canals 

Concreting in Cold Weather 

That Alley of Yours 

Concrete Rouses and Why to Build Them 



Are booklets of distinct educational value, 
free from advertising, and in most cases 
will serve as supplemental classroom texts. 
If interested in any of these publications, 
write the 



Portland Cement Association 

111 West Washington Street 
CHICAGO 



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